算法找到给定的电弧，其中心，半径和角度3分

Question

鉴于3点A，B和C

如何找到并开始于甲弧，中C端和通过B;其中心的坐标，r和r'的半径和角度？

Answer 1

有几种方法去做这个。这是一种算法：

1. 让您COORDS

   A = {XA，YA}

   B = {XB，YB}

   C = {XC，YC}

   d = {xd，yd}

2. 计算AB线和BC线的中点

   mid_AB = {（XA + XB）/ 2，（YA + YB）/ 2}

   mid_BC = {（的xB + XC）/ 2，（YB + YC）/ 2}

3. 找到线AB和BC

   的斜坡

   slope_AB =（YB-YA）/（XB-XA）

   slope_BC =（YC-YB）/（XC-XB）

通过运行

4. 构建线中点PERPENDICULAR AB和BC（谢谢Yves捕捉负）

   Slope_perp_AB！= - （slope_AB）^（ - 1）

   Slope_perp_BC = - （slope_BC）^（ - 1）

   ***与Slope_perp_AB线通过mid_AB运行

   ***与Slope_perp_BC线通过mid_BC

5. 运行设置两个方程彼此相等，解决找到交集！ 这给你点d = {xd，yd} !!!

计算半径和角度现在微不足道的中心点d！

Answer 2

步骤1

查找AB和BC的垂直平分线。

第2步

找到这些线相交的点。

你会发现点将是你想要的圆的中心。

步骤3

计算从您在步骤2中发现中间的三个点，这将是半径的圆你一个的距离。

注意点A，B和C不得在同一行。在执行步骤1至3之前，必须先检查该步骤。

Answer 3

对此的解决方案与“非超定系统的最佳拟合圈”几乎相同。由于您有三个点，它们正好位于以圆心(0,0)（给定）为圆心的圆弧上，所以系统可以精确求解，而不需要最小二乘法逼近。

Finding the Center of a Circle Given 3 Points 


Date: 05/25/2000 at 00:14:35 
From: Alison Jaworski 
Subject: finding the coordinates of the center of a circle 

Hi, 

Can you help me? If I have the x and y coordinates of 3 points - i.e. 
(x1,y1), (x2,y2) and (x3,y3) - how do I find the coordinates of the 
center of a circle on whose circumference the points lie? 

Thank you. 


Date: 05/25/2000 at 10:45:58 
From: Doctor Rob 
Subject: Re: finding the coordinates of the center of a circle 

Thanks for writing to Ask Dr. Math, Alison. 

Let (h,k) be the coordinates of the center of the circle, and r its 
radius. Then the equation of the circle is: 

    (x-h)^2 + (y-k)^2 = r^2 

Since the three points all lie on the circle, their coordinates will 
satisfy this equation. That gives you three equations: 

    (x1-h)^2 + (y1-k)^2 = r^2 
    (x2-h)^2 + (y2-k)^2 = r^2 
    (x3-h)^2 + (y3-k)^2 = r^2 

in the three unknowns h, k, and r. To solve these, subtract the first 
from the other two. That will eliminate r, h^2, and k^2 from the last 
two equations, leaving you with two simultaneous linear equations in 
the two unknowns h and k. Solve these, and you'll have the coordinates 
(h,k) of the center of the circle. Finally, set: 

    r = sqrt[(x1-h)^2+(y1-k)^2] 

and you'll have everything you need to know about the circle. 

This can all be done symbolically, of course, but you'll get some 
pretty complicated expressions for h and k. The simplest forms of 
these involve determinants, if you know what they are: 

     |x1^2+y1^2 y1 1|  |x1 x1^2+y1^2 1| 
     |x2^2+y2^2 y2 1|  |x2 x2^2+y2^2 1| 
     |x3^2+y3^2 y3 1|  |x3 x3^2+y3^2 1| 
    h = ------------------, k = ------------------ 
      |x1 y1 1|    |x1 y1 1| 
      2*|x2 y2 1|    2*|x2 y2 1| 
      |x3 y3 1|    |x3 y3 1| 

Example: Suppose a circle passes through the points (4,1), (-3,7), and 
(5,-2). Then we know that: 

    (h-4)^2 + (k-1)^2 = r^2 
    (h+3)^2 + (k-7)^2 = r^2 
    (h-5)^2 + (k+2)^2 = r^2 

Subtracting the first from the other two, you get: 

    (h+3)^2 - (h-4)^2 + (k-7)^2 - (k-1)^2 = 0 
    (h-5)^2 - (h-4)^2 + (k+2)^2 - (k-1)^2 = 0 

    h^2+6*h+9 - h^2+8*h-16 + k^2-14*k+49 - k^2+2*k-1 = 0 
    h^2-10*h+25 - h^2+8*h-16 + k^2+4*k+4 - k^2+2*k-1 = 0 

    14*h - 12*k + 41 = 0 
    -2*h + 6*k + 12 = 0 

    10*h + 65 = 0 
    30*k + 125 = 0 

    h = -13/2 
    k = -25/6 

Then 

    r = sqrt[(4+13/2)^2 + (1+25/6)^2] 
     = sqrt[4930]/6 

Thus the equation of the circle is: 

    (x+13/2)^2 + (y+25/6)^2 = 4930/36 

- Doctor Rob, The Math Forum 
    http://mathforum.org/dr.math/ 

参考

---

1. 寻找鉴于三点一个圈，访问的2014年4月1日的中心，<http://mathforum.org/library/drmath/view/55239.html>

Answer 4

你有三个方程来确定三个未知XM，YM和R，

(xA-xM)^2+(yA-yM^2) = R^2 

等从B中减去甲方程和C方程给出

2*(xB-xA)*xM+2*(yB-yA)*yM = xB^2-xA^2+yB^2-yA^2 
2*(xC-xA)*xM+2*(yC-yA)*yM = xC^2-xA^2+yC^2-yA^2 

通过求解这2×2线性系统，您可以获得圆的中心点，插入任何原始方程都可得出半径。

Answer 5

圆的中心是等距离的三个已知点：

(X-Xa)^2+(Y-Ya)^2 = (X-Xb)^2+(Y-Yb)^2 = (X-Xc)^2+(Y-Yc)^2 

减去来自所述第二与第三所述第一构件，我们重新组合后得到：

2(Xa-Xb) X + 2(Ya-Yb) Y + Xb^2+Yb^2-Xa^2-Ya^2 = 0 
2(Xa-Xc) X + 2(Ya-Yc) Y + Xc^2+Yc^2-Xa^2-Ya^2 = 0 

该线性系统在两个未知数中的两个方程很容易用克莱默法则解决。

半径和角度可以通过围绕中心笛卡尔到极变换中找到：

R= Sqrt((Xa-X)^2+(Ya-Y)^2) 

Ta= atan2(Ya-Y, Xa-X) 
Tc= atan2(Yc-Y, Xc-X) 

但是你还是会错过一两件事：什么是弧形的相关部分？小于或大于半圈？从Ta到Tb或从Tb到2 Pi到Ta + 2 Pi，还是什么？答案似乎不像看起来那么明显，试试吧（因为三个角度Ta,Tb和Tc都未定义为2 Pi的倍数，您无法对它们进行排序）！

提示：考虑三角形ABC区域的正负号，正好是系统行列式的一半。它会告诉你B是在AC的左边还是右边。

Answer 6

有一个鲜为人知的结果通过3点给人一种圆的隐式方程：

|Z X Y 1| 
|Za Xa Ya 1| 
|Zb Xb Yb 1| = 0 
|Zc Xc Yc 1| 

，我们已经为简明起见定义Z:= X^2 + Y^2。

计算的3×3的未成年人，我们发展成：

M00 Z + M10 X + M20 Y + M30 = 0 

和，归一化后，我们得到了通常的二度方程：

X^2 + Y^2 + 2U X + 2V Y + W = 0 

这可以被改写为：

(X - U)^2 + (Y - V)^2 = U^2 + V^2 - W 

立即给中心(U, V) = (-M10/2.M00, -M20/2.M00)和半径R^2 = U^2 + V^2 - M30/M00。